Solve the third-order initial value problem below using the method of Laplace transforms y'''-2y"-21y'-18y=-18 y(0)=2 y'(0)=7 y"(0)=95

Solve the third-order initial value problem below using the method of Laplace transforms y'''-2y"-21y'-18y=-18 y(0)=2 y'(0)=7 y"(0)=95

Question
Laplace transform
asked 2020-11-09
Solve the third-order initial value problem below using the method of Laplace transforms
\(y'''-2y"-21y'-18y=-18\)
\(y(0)=2\)
\(y'(0)=7\)
\(y"(0)=95\)

Answers (1)

2020-11-10
Step 1
Taking Laplace Transform in both direction:
\(L\left\{y'''-2y"-21y'-18y\right\}=L\left\{18\right\}\)
\(L\left\{F^n(t)\right\}=s^nY(s)-s^2y(0)-sy'(0)-y"(0)\)
\(L\left\{y'''\right\}=s^3Y(s)-s^2y(0)-sy'(0)-y'(0)\)
\(L\left\{y"\right\}=s^2Y(s)-sy(0)-y'(0)\)
\(L\left\{y'\right\}=sY(s)-y(0)\)
\(L\left\{y\right\}=Y(s)\)
\(\Rightarrow \left[s^3Y(s)-s^2y(0)-sy'(0)-y'(0)\right]-2\left[s^2Y(s)-sy(0)-y'(0)\right]-21\left[sY(s)-y(0)\right]-18\left[Y(s)\right]=L\left\{-18\right\}\)
\(\Rightarrow [s^3Y(s)-s^2(2)-7s-95]-2[s^2Y(s)-s(2)-7]-21[sY(s)-2]-18[Y(s)]=-\frac{18}{s}\)
\(\Rightarrow {s^3-2s^2-21s-18}Y(s)-2s^2-7s-95+4s+14+42=-\frac{18}{s}\)
\(\Rightarrow Y(s)=\frac{2s^2+3s-\frac{18}{s}+39}{s^3-2s^2-21s-18}=\frac{2s^3+3s^2-18+39s}{s(s+1)(s+3)(s-6)}\)
Step 2
find the factor:
\(Y(s)=\frac{A}{s} + \frac{B}{(s+1)} + \frac{C}{(s+3)} +\frac{D}{(s-6)}\)
\(Y(s)=\frac{2s^3+3s^2-18+39s}{s(s+1)(s+3)(s-6)}\)
\(\frac{A}{s} + \frac{B}{(s+1)} + \frac{C}{(s+3)} +\frac{D}{(s-6)} = \frac{\frac{2(0)^3+3(0)^2-18+39(0)}{(1)(3)(-6)}}{s} +\frac{\frac{2(-1)^3+3(-1)^2-18+39(-1)}{(-1)(-1+3)(-1-6)}}{(s+1)} + \frac{\frac{2(-3)^3+3(-3)^2-18+39(-3)}{(-3)(-3+1)(-3-6)}}{(s+3)} + \frac{\frac{2(6)^3+3(6)^2-18+39(6)}{(6)(7)(9)}}{(s-6)}\)
\(=\frac{1}{s} +\frac{(-4)}{(s+1)} + \frac{3}{(s+3)} +\frac{2}{(s-6)}\)
Step 3
taking the inverse Laplace transform of the function as find y(t):
\(y(t)=L^{-1}\left\{\frac{1}{s} +\frac{(-4)}{(s+1)} + \frac{3}{(s+3)} +\frac{2}{(s-6)}\right\}\)
\(\text{Using } L^{-1}\left(\frac{1}{(s-a)}\right)=e^{at}\)
\(=1+-4e^{-t}+3e^{-3t}+2e^{6t}\)
0

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